Passing a 2D pointer array to a function with a 1D pointer array parameter

I am currently doing a project where I am going to pass a 2D pointer array to a function. Here is the function implementation:

void
affiche_Tab2D(int *ptr, int n, int m)
{
    if (n>0 && m>0 && ptr!=NULL)
    {
        int (*lignePtr)[m]; // <-- Its my first time to see this declaration

        lignePtr = (int (*)[m]) ptr; // <-- and this too....

        for (int i = 0 ; i < n ; i++)
        {
            for (int j = 0 ; j < m ; j++) 
            {
                printf("%5d ",lignePtr[i][j]);
            }
            printf("\b\n");
        }
    }
}

Notice that ptr is 2D array but it uses a single pointer. Before, I used to pass 2D array using double pointers. In my main I have set up a 2D array (double pointer) and finding ways how to send it to that function.

Here are the function calls I tried which does not work:

int 
main(int argc, char * argv[]) 
{
    int ** numbers_table;
    int i, j;

    numbers_table = (int **)malloc(10 * sizeof(int *));

    for(i = 0; i < 10; i++)
        numbers_table[i] = (int *)malloc(10 * sizeof(int));

    for(i = 0; i < 10; i++)
        for(j = 0; j < 10; j++)
            numbers_table[i][j] = i + j;

    // Failed function call 1
    // affiche_Tab2D(numbers_table, 10, 10);

    // Failed function call 2
    // affiche_Tab2D(&numbers_table, 10, 10);

    for(i = 0; i < 10; i++)
    free(numbers_table[i]);

    free(numbers_table);

    return 0;
}